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I want to show that, when $\mu=\mu_0$, then gamma family $\Gamma(a,b)$ is a conjugate prior to inverse Gaussian with density $f(x,\mu,\lambda)=\sqrt{\frac{\lambda}{2\pi x^2}}exp[-\frac{\lambda(x-\mu)^2}{2\mu^2x}]$.

I was able to verify that $IG$ is an exponential family generated by $\bf{T}$$(X)=-1/2(X,X^{-1})^T$ and $h(x)=\sqrt{\frac{1}{2\pi x^3}}$ but I am pretty confused on how to show the conjugate prior.

Any helps would be deeply appreciated!!

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    $\begingroup$ The conjugate prior for an exponential family $f(x;\theta)=h(x)\exp\{\theta T(x)-\psi(\theta)\}$ is $\pi(\lambda;a,b)\propto\exp\{\theta a-b\psi(\theta)\}$, the measure $h(\cdot)$ does not matter. $\endgroup$ – Xi'an Mar 9 at 7:10
  • $\begingroup$ When $\mu=\mu_0$, a fixed value, the sufficient statistic is no longer $(X,X^{-1})$. $\endgroup$ – Xi'an Mar 9 at 7:12
  • $\begingroup$ @Xi'an Yeah, I know the formula. I have trouble deriving it. $\endgroup$ – statwoman Mar 9 at 14:54
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$$\pi_t(\theta)=exp(\Sigma_{j=1}^k\eta_j(\boldsymbol{\theta})t_j-t_{k+1}B(\boldsymbol{\theta})-logw(\mathbf{t}))$$ $$w(t)=\int...\int\int_Rexp(\Sigma t_j\eta_j-t_{k+1}B)d\theta_1...d\theta_k$$ $$B(\theta)=-\frac{\lambda}{\mu}-\frac{log\lambda}{2}$$ Since $\mu$ is known then k=1 in this part. taking $a=t1$ and $b=\frac{t2-t1}{\mu}$, we yield: $$w(t)=\int_0^\infty exp(-b\lambda)\lambda^{a/2}d\lambda=b^{-\frac{a+1}{2}}\Gamma(\frac{a+1}{2})$$ $$\Pi_t(\theta)=\frac{\lambda^{\frac{a+1}{2}}e^{-b\lambda}}{\Gamma(\frac{a+1}{2})}b^{\frac{a+1}{2}}$$ We can conclude that this shows the gamma distribution with parameters $\alpha=\frac{a+1}{2}$ and $\beta=b$.

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